2. The voltage of an electrochemical cell depends on the redox reaction occurring in the cell. Given a redox reaction below:

Zn(s) + Pb2+(aq) --> Zn2+(aq) + Pb(s)
Pb2+(aq) + 2 e--->Pb(s)Eº = -0.13V
Zn2+(aq) + 2 e--->Zn(s)Eº = -0.76V
Calculate the E.M.F (Voltage) of the cell

To calculate the electromotive force (EMF) or voltage of the electrochemical cell, you need to use the standard reduction potentials (Eº) of the half-reactions involved in the reaction.

In this case, the given half-reactions are:
1. Pb2+(aq) + 2 e- → Pb(s) with Eº = -0.13V
2. Zn2+(aq) + 2 e- → Zn(s) with Eº = -0.76V

To determine the overall reaction and calculate the EMF, you need to refer to the standard reduction potentials table, which provides the standard reduction potentials for different half-cells.

The EMF or voltage of an electrochemical cell can be calculated using the formula:

EMF of the cell = Eº reduction of the cathode - Eº reduction of the anode

In the given reaction, the cathode is the reduction half-reaction of Pb2+, and the anode is the oxidation half-reaction of Zn. Therefore, the reduction potential of the cathode (Eº reduction of the cathode) is -0.13V, and the reduction potential of the anode (Eº reduction of the anode) is -0.76V.

Let's calculate the EMF of the cell using the above formula:

EMF of the cell = Eº reduction of the cathode - Eº reduction of the anode
= -0.13V - (-0.76V)
= -0.13V + 0.76V
= 0.63V

Therefore, the EMF or voltage of the cell is 0.63 volts.